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I'm reaing the Griffiths & Harris "Principles of algebraic geometry", and I'm blocked at page 80. Previously it has defined the hermitian metric $ds^2$ on a complex manifold $M$ as $$(\ ,\ )_z:T'_z(M)\otimes\bar{T'_z(M)}\longrightarrow\mathbb{C}$$ depending smoothly on z, and it has chose an unitary coframe, i.e. an n-tuple of (1,0)-forms $(\varphi_1,...,\varphi_n)$ such that $(\varphi_1(z),...,\varphi_n(z))$ is an orthonormal base for $T^{\ast'}_z(M)$ (respect to the induced ( , )). It says that $ds^2$ induces an hermitian metric on the tensor bundle $T^{\ast(p,q)}(M)$ given by taking the basis $(\varphi_I(z)\wedge\varphi_J(z))_{cardI=p, cardJ=q}$ to be orthogonal and of length $2^{p+q}$. I don't understand why is not the lenght 1?? If $cardI=1$ and $cardJ=0$ than $||\varphi_i(z)||^2=(\varphi_i(z),\varphi_i(z))=1$ because the base in ortonormal. Where is my mistake?

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    You're confusing a few different things. (1) GH are *defining* a new metric on the space of $(p,q)$-forms for any $p, q$. Even for $(p,q) = (1,0)$ this is a metric on the *dual* space to the tangent bundle, and thus not the same as the original metric. (2) When they say "length", they mean the length of a vector, not length of the index sets $I$ and $J$. – 2017-02-20
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    Thank you, now it's clearer. – 2017-03-09

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